If `(x^(2))/(x-2)>0` then `xin`__________

Answer

To solve the inequality \(\frac{x^2}{x-2} > 0\), we will follow these steps:

### Step 1: Identify the critical points
The expression \(\frac{x^2}{x-2}\) will be positive when both the numerator and denominator have the same sign (both positive or both negative).

1. **Numerator**: \(x^2 = 0\) gives the critical point \(x = 0\).
2. **Denominator**: \(x - 2 = 0\) gives the critical point \(x = 2\).

### Step 2: Determine the intervals
The critical points divide the number line into intervals. We will analyze the sign of the expression in the following intervals:
- \( (-\infty, 0) \)
- \( (0, 2) \)
- \( (2, \infty) \)

### Step 3: Test each interval
We will choose a test point from each interval to determine the sign of \(\frac{x^2}{x-2}\).

1. **Interval \( (-\infty, 0) \)**: Choose \(x = -1\)
\[ \frac{(-1)^2}{-1 - 2} = \frac{1}{-3} < 0 \quad \text{(negative)} \]

2. **Interval \( (0, 2) \)**: Choose \(x = 1\)
\[ \frac{1^2}{1 - 2} = \frac{1}{-1} < 0 \quad \text{(negative)} \]

3. **Interval \( (2, \infty) \)**: Choose \(x = 3\)
\[ \frac{3^2}{3 - 2} = \frac{9}{1} > 0 \quad \text{(positive)} \]

### Step 4: Combine the results
From our testing:
- The expression is negative in the intervals \((- \infty, 0)\) and \((0, 2)\).
- The expression is positive in the interval \((2, \infty)\).

### Step 5: Write the solution
Since we are looking for where the expression is greater than zero, the solution is:
\[ x \in (2, \infty) \]

### Final Answer
Thus, \(x \in (2, \infty)\).

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